Updated: January 26, 2016
Calculus III
Section 14.7
Math 232
Calculus III
Brian Veitch •
Fall 2015 •
Northern Illinois University
14.7 Maximum and Minimum Values
Definition 1: Local Extrema
f (x, y) has a local minimum at (a, b) if f (x, y) ≥ f (a, b) when (x, y) is near (a, b).
f (x, y) has a local maximum at (a, b) if f (x, y) ≤ f (a, b) when (x, y) is near (a, b).
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Updated: January 26, 2016
Calculus III
Section 14.7
Theorem 1
If f has a local maximum or minimum at (a, b) and fx and fy exist, then fx (a, b) = 0
AND fy (a, b) = 0
Like in calculus I, it means all potential max/mins must occur when fx (a, b) = 0
and fy (a, b) = 0. Call these Critical Points.
Example 1
Let f (x, y) = x2 + y 2 − 2x − 6y + 14. Find all critical points.
1. Start by finding fx and fy .
fx = 2x − 2
fy = 2y − 6
2. Find all points (x, y) where fx = 0 and fy = 0.
fx = 2x − 2 = 0 when x = 1
fy = 2y − 6 = 0 when y = 3
3. Since there are no contradictions between our solutions, we have a critical point at
(1, 3).
In Calculus I we talked about how not all critical points are local extrema. We can extend
that idea to the three-dimensional coordinate system. Consider the graph below.
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Updated: January 26, 2016
Calculus III
Section 14.7
Definition 2: Saddle Point at (a, b)
Consider P (a, b) at the origin.
If you
travel the curve along the y-axis (in either direction) the slope is positive. As
you travel along the x-axis (in either direction) the slope is negative. The point
can neither be a local maximum nor a local minimum.
We call the point (a, b) a saddle point.
Definition 3: Second Derivative Test
Assume the second partial derivatives of f are continuous on a disk with center (a, b).
Suppose fx (a, b) = 0, fy (a, b) = 0 and
D = D(a, b) = fxx (a, b)fyy (a, b) − [fxy (a, b)]2
fxx fxy D=
fyx fyy 1. If D > 0 and fxx (a, b) > 0 then f (a, b) is a local minimum.
2. If D > 0 and fxx (a, b) < 0 then f (a, b) is a local maximum.
3. If D < 0, then f (a, b) is neither a maximum nor minimum (Saddle Point)
4. If D = 0, the test is inconclusive.
Example 2
Find the local extrema and saddle points of f (x, y) = x4 + y 4 − 4xy + 1.
1. Let’s start by finding all the critical points.
(a) fx = 4x3 − 4y, fy = 4y 3 − 4x. We must solve
fx = 4x3 − 4y = 0
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Updated: January 26, 2016
Calculus III
Section 14.7
fy = 4y 3 − 4x = 0
Solving the first equation for y we get y = x3 . Plug this into the second equation
and we get
4(x3 )3 − 4x = 0
4x9 − 4x = 0
4x(x8 − 1) = 0
4x(x4 − 1)(x4 + 1) = 0
4x(x2 − 1)(x2 + 1)(x4 + 1) = 0
4x(x − 1)(x + 1)(x2 + 1)(x4 + 1) = 0
x = 0, 1, −1
(b) For each x-value we need its corresponding y-value.
If x = 0, y = (0)3 = 0
If x = −1, y = (−1)3 = −1
If x = 1, y = (1)3 = 1
(c) We have three critical points: (0, 0), (1, 1), and (−1, −1).
2. For each of the critical points we need to find D(x, y). To do this we need fxx , fyy ,
and fxy .
fxx = 12x2
fyy = 12y 2
fxy = −4
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Updated: January 26, 2016
Calculus III
Section 14.7
(a) Critical Point (0, 0):
D(0, 0) = fxx (0, 0)fyy (0, 0) − [fxy (0, 0)]2
= 12(0)2 · 12(0)2 − [−4]2
= −16
Since D < 0, the critical point (0, 0, 1) is a saddle point.
(b) Critical Point (1, 1):
D(1, 1) = fxx (1, 1)fyy (1, 1) − [fxy (1, 1)]2
= 12(1)2 · 12(1)2 − [−4]2
128
Since D > 0 and fxx (1, 1) > 0, the critical point (1, 1, −1) is a local minimum.
(c) Critical Point (−1, −1):
D(−1, −1) = fxx (−1, −1)fyy (−1, −1) − [fxy (−1, −1)]2
= 12(−1)2 · 12(−1)2 − [−4]2
128
Since D > 0 and fxx (−1, −1) > 0, the critical point (−1, −1, −1) is a local
minimum.
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Updated: January 26, 2016
Calculus III
Section 14.7
Example 3
Find the shortest distance from (1, 0, −2) to the plane x + 2y + z = 4
The distance between two points (x, y, z) and (1, 0, −2) in a three-dimensional coordinates
system is
d=
p
(x − 1)2 + (y − 0)2 + (z + 2)2
One problem we have is the function d(x, y) has a z in it. We need to write z in term of x
and y. Since we know the point must be on the plane z = 4 − x − 2y we can substitute this
in for z
p
(x − 1)2 + y 2 + ((4 − x − 2y) + 2)
p
d = (x − 1)2 + y 2 + (6 − x − 2y)2
d=
1. Find dx and dy
2(x − 1) + 2(6 − x − 2y) · (−1)
dx = p
2 (x − 1)2 + y 2 + (6 − x − 2y)2
−14 + 4x + 4y
dx = p
2 (x − 1)2 + y 2 + (6 − x − 2y)2
2y + 2(6 − x − 2y) · (−2)
dy = p
2 (x − 1)2 + y 2 + (6 − x − 2y)2
−24 + 4x + 10y
dy = p
2 (x − 1)2 + y 2 + (6 − x − 2y)2
2. Solve dx = 0 and dy = 0
dx = 0 ⇒ 4x + 4y − 14 = 0
dy = 0 ⇒ 4x + 10y − 24 = 0
Solving these two equations we get y =
5
11
and x =
3
6
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Updated: January 26, 2016
Calculus III
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Section 14.7